In $\textbf{Z}[\sqrt{30}]$, the number $5$ splits, since, for example, $N(5 + \sqrt{30}) = -5$. But the ideal $\langle 5 \rangle$ is a ramifying ideal, since it is equal to $\langle 5, \sqrt{30} \rangle^2$. It should have been immediately obvious to me (but it wasn't) that $(5 + \sqrt{30}) \in \langle 5, \sqrt{30} \rangle$.
Given the factorization of $30$, it is clear that $\langle 2 \rangle$ and $\langle 3 \rangle$ are also ramifying ideals. By examining the least significant digits of the squares, it is also clear that, as numbers, $2$ and $3$ are inert and therefore irreducible.
But by what other way can I quickly verify that $2$ and $3$ are inert but generate ramifying ideals? And can that other way be readily carried over to a domain like $\textbf{Z}[\sqrt{42}]$, where the least significant digits of the squares might not necessarily help in discerning primes that split but generate ramifying ideals from those that are inert but also generate ramifying ideals?
EDIT: I had a stray exponent $2$ at first. Sorry for the confusion.